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This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(\{e\})$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

Remark: This question was motivated by a brief discussion with Stefaan Vaes.

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(\{e\})$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(\{e\})$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(\{e\})$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

Remark: This question was motivated by a brief discussion with Stefaan Vaes.

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(e)$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveals problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

This post is freely inspired by the basic rules of Go (game), usually played on a $19 \times 19$ grid graph.

Consider the $\mathbb{Z}^2$ grid. We can assign to each vertex a state "black" ($b$), "white" ($w$) or "empty" ($e$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $\phi^{-1}(\{b\})$ and $\phi^{-1}(\{w\})$ decompose into connected components $c$. Let $d(c)$ be the number of liberties of $c$, i.e. the number of elements of $\phi^{-1}(\{e\})$ connected to $c$ by an edge.
The following picture on the left shows a white connected component $c_1$ with number of liberties $d(c_1)=8$ and a black connected component $c_2$ with $d(c_2) = 5$. On the right, there is a black connected component $c_3$ with $d(c_3)=0$.

$\hspace{2cm}$

A state $\phi$ is finite if $|\phi^{-1}(\{b,w\})| < \infty$, and is admissible if all its connected components $c$ have at least one liberty. Let $\mathcal{F}$ be the set of finite admissible states. Let $g \in \mathbb{Z}^2$ and consider the map $B_g: \mathcal{F} \to \mathcal{F}$ such that if $\phi(g) \neq e$ then $B_g(\phi) = \phi$, else $B_g(\phi)$ is the state made by first putting the vertex $g$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$\hspace{0.7cm}$ $\hspace{0.7cm}$

The map $W_g$ is defined similarly by interchanging black and white. Now, let $H$ be the Hilbert space $\ell^2(\mathcal{F})$. We extend $B_g$ and $W_g$ by linearity on $H$. The adjoint $B^*_g$ of $B_g$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $W^*_g$ of $W_g$ is given by a similar formula.

Let $\mathcal{A}$ be the $*$-algebra generated by the set $\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$.

Question: Is $H$ an irreducible representation of $\mathcal{A}$?

If yes, this means that the von Neumann algebra $M:=\mathcal{A}''$ equals $B(H)$, the algebra of all bounded operators. Else, what is $M$?

Remark: these questions can be extended to any finitely generated group $\Gamma$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $g$, $\Vert B_g \Vert^2$ should "intuitively" be $2^d$ (with $d$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $k$-regular (with $k$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.